What is the vertex of the graph of f(x) = x2 + 10x - 9? a) (-5, -34) b) (-5, -9) c) (5, -9) d) (5, 66)

A) (-5, -34)

Step-by-step explanation:

f(x) = x^2 + 10x - 9

We complete the square to get the equation in vertex form

Take the coefficient of the x term and divide by 2 then square it.  We add it and then subtract it not to change the value of the equation

f(x) = x^2 + 10x +(10/2)^2 - (10/2)^2 - 9

f(x) = x^2 +10x +25 -25 -9

f(x) = (x^2 +10x +25) -34

The term in parentheses simplified to (x+10/2) ^2

     = (x+5)^2 -34

   = (x - -5)^2 -34

This is in the form (x-h)^2 +k

The vertex is (h,k)  h=-5 and k=-34

(-5,-34)

the length of o'p' is .

step-by-step explanation:

it is given that the quadrilateral m'n'o'p' is the image of quadrilateral mnop after dilation.

the center of dilation is origin and the scale factor is 1.5.

in dilation the image and preimge are similar figures. it means the sides of image and preimage are proportional.

the scale factor is defined as

the scale factor is 1.5.

multiply both sides by op.

note: if the value of op is given, then multiply the value of op with 1.5 to get the length of side o'p'.

therefore length of o'p' is .

won't that be plagrisim thought?

step-by-step explanation: